# Is a ring a set?

We know that a ring consists of a set equipped with two binary operations. My question is whether a ring is a set or not. For example, we can have $(\mathbb{R},+,-)$ where $\mathbb{R}$ is a set and $+$ and $-$ are binary operations associated with the set. Note that binary operations are functions, and functions are set, so we have a 3-tuple consisting of three sets. My first question is whether this tuple itself is a set? i.e. what exactly is a tuple?

In addition, the problem is I am not comfortable with defining ring as something with soemthing else. What exactly does it mean by "with"? (for example, is it a union?) it just seems overly informal.

Any help is apprecaited.

• @janmarqz Because the OP seems to be wondering "what does 'with' mean?" Mar 17, 2016 at 3:47
• It is better not to accept answers so fast, since this effectively locks out people who only visit the site once a day. Mar 17, 2016 at 5:49
• @goblin: if you let an accepted answer prevent you from answering a question, you are doing it wrong & entirely missing the point of the SE network. Mar 17, 2016 at 13:16
• @KyleKanos, but I don't let it prevent me from answering. I'm just saying, it's better to wait. Btw, I don't think the SE network has a particularly good model. Its better than anything else out there, but it isn't good. Mar 17, 2016 at 13:17
• Instead of saying a ring is a set with two operations blah just say that a ring is a triple where the first element is a set and the other two are operations over the first set. This is what is also used when speaking, e.g., about automatas in theoretical computer science. An automaton is a 5-uple $(Q, \Sigma, \delta, q_0, F)$ where $A$ is a set (of states), $\Sigma$ is a set (the alphabet), $\delta: Q \times \Sigma \to Q$ is a function, $q_0 \in Q$ is the initial state and $F \subseteq Q$ are the final states. You could say that an automaton is a set $Q$ with a few associated functions... Mar 17, 2016 at 17:37

You only need to see the formality once to never want to see it again.

The ordered pair $(a,b)$ is defined to be, as a set, $\{\{a\},\{a,b\}\}$. So we could say an ordered triple $(a,b,c)$ is an ordered pair $$((a,b),c)=\{\{\{\{a\},\{a,b\}\}\},\{\{\{a\},\{a,b\}\},c\}\}$$ Satisfying ourselves that such a thing is existentially valid we can freely write $(a,b,c)$ to mean the same thing with less clunky notation.

• +1 just for "You only need to see the formality once to never want to see it again." - so true in so many places. Mar 17, 2016 at 5:39
• When defined as a set, $(a,b)$ is usually defined to be $\{\{a\}, \{a,b\}\}$.
– MJD
Mar 17, 2016 at 13:52
• @MJD thanks for the correction. That makes it even uglier. Mar 17, 2016 at 14:00
• I would nitpick to say that the ordered pair is not necessarily defined this way, but we implement it in set theory in this way. We all know what an ordered pair is, and we need to represent it in set theory; this is how we implement it. It's not really a definition of an ordered pair, so much as an implementation. Mar 18, 2016 at 11:19
• I would go further and say that the set-theoretic implementation of a pair only satisfies me that set theory can implement pairs, rather than that pairs are "existentially valid". It doesn't seem to me that the existence of sets is any more obvious than the existence of pairs. If you told me that sets couldn't implement pairs, I definitely wouldn't conclude that pairs therefore don't exist. Mar 19, 2016 at 17:40

This will make it formal for you. Let $S$ be any set. Then the ring $R$ is any element of the set $$R=(S,f,g)\in \{S\}\times S^{S\times S}\times S^{S\times S},$$ where $f$ and $g$ are functions, elements of $S^{S\times S}$ satisfying: $$f(f(a,b),c)=f(a,f(b,c)),$$ $$f(a,b)=f(b,a),$$ $$\exists e\in S\text{ such that } f(a,e)=a\;\forall a\in S,$$ $$\forall a\in S, \exists b\in S\text{ such that } f(a,b)=e,$$ $$g(g(a,b),c)=g(a,g(b,c)),$$ $$g(a,f(b,c))=f(g(a,b),g(a,c)).$$

Now do yourself a favor, and don't always treat rings with such formality.

• So then that wouldn't make it a set right? Unless it's a singleton set $\{(S,f,g)\}$ Mar 19, 2016 at 15:54
• Just because it is an element of a set doesn't mean that it is not also a set (Matt Samuel shows how this works in the accepted answer). I think that this is a more parsimonious way of rigorously defining a ring. Mar 19, 2016 at 15:56
• Oh yes I agree, I mean elements of a power set are sets. I just mean from this answer it's not obvious to me that it's a set, unless you make it so like in Matt Samuels answer. Also it wouldn't make any difference to how I would use a ring in either case but I could imagine the distinction might matter in some way? Mar 19, 2016 at 15:59
• Actually I get it now, thanks! I think I'd need to see how Cartesian products are defined rigorously to piece the two answers together Mar 19, 2016 at 16:03

I'll try to address both your questions from the viewpoint of category theory.

Is a ring a set?

No.* But we can always treat a ring $R$ as if it were a set, in the following way; there's a functor $$U:\mathbf{Ring} \rightarrow \mathbf{Set}$$ given on objects by $U(S,+,\times) = S$. This is called the underlying set functor (or "forgetful functor to $\mathbf{Set}$") and it allows us to treat rings as if they were sets and morphisms of rings as if they were functions. This in turn allows us to "pull back" structure on $\mathbf{Set}$ to get structure on $\mathbf{Ring}$.

For example, there's a notion of finiteness for sets. Hence we can define that ring $R$ is finite iff the set $U(R)$ is finite. So we've "pulled back" the notion of finiteness across $U$. Similarly, there's a notion of surjectivity for functions. Hence we can define that a ring homomorphism $f : R_0 \rightarrow R_1$ is surjective iff the corresponding underlying function $U(f) : U(R_0) \rightarrow U(R_1)$ is surjective. Again, this pulls surjectivity back across $U$.

*Except in material set theory, in which it is typically assumed that everything is a set, even things like ordered pairs. This doesn't have too much bearing on everyday mathematics, though.

What does "with" mean?

This is a much, much harder question to answer in a satisfactory way; indeed, basic category theory doesn't even attempt to give this question an answer. But if you have some familiarity with double categories, we can indeed give this question an answer; in particular, see Susan Niefield's article here on the gluing construction (but only once you're ready.)

• I think this is also a great answer from a different perspective. I would have accepted this answer if SE allowed two accepted answers. Thanks!
– Kun
Mar 17, 2016 at 16:42
• A comment on surjectivity, only relevant to advanced readers: "surjectivity" is not a categorical notion; the corresponding categorical notion is an "epimorphism". In almost all familiar contexts, epimorphisms end up being the same as surjections. But in the category of rings they're DIFFERENT. In fact, the natural inclusion $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is an epimorphism, even though it's definitely not a surjection. Mar 18, 2016 at 12:39
• @TomChurch, yes, of course. That's why its easiest to just "pull it back", as I describe, using the fact we're working concrete over $\mathbf{Set}$. On the other hand, note that the regular epimorphisms in $\mathbf{Ring}$ are precisely the surjective homomorphisms. Mar 18, 2016 at 12:41
• It actually turns out there are good reasons to think of $\mathbb{Z}\hookrightarrow\mathbb{Q}$ as just as "surjection-y" as $\mathbb{Z}\twoheadrightarrow\mathbb{Z}/p\mathbb{Z}$, at least in some contexts (when we're thinking of these rings as functions on some space). Mar 18, 2016 at 12:43
• Possibly a silly question, because I've never studied category theory, but how is the category $Ring$ defined if you can't take it as a set and define it the usual way? Mar 19, 2016 at 15:46

You could say that the term "ring" simultaneously refers to two related things.

The first is that the term ring refers to a certain amount of mathematical data. This data includes a set $R$, as well as two functions $R \times R \to R$ with certain properties. The second is that the term ring is also used to refer to the set $R$ itself.

We often interchange these uses. For example, it is considered perfectly valid to "take an element of $R$," as if the ring is just a set, but it also makes sense to say "Let $R$ be a ring," where we understand that we have the data of a set as well as the corresponding addition and multiplication.

This idea is not unique to the term "ring." It applies to nearly every mathematical term defined with any structure. Terms like "group", "manifold", and many others can be used to simultaneously refer to the combined data of set and structure, but also refer to the set itself.

• +1. Another very common example is "space" (as in "metric space" or "topological space"). Mar 18, 2016 at 1:24

When we say a set $X$ is equipped with two binary operations $P$ and $Q$, the word "with" doesn't really signify anything by itself. It's the use of the words "equipped with" and "and" in this pattern (in the context of defining an algebraic object such as a ring) that gives us a way to express the idea of the ordered triple $(X,P,Q)$ in a way that's easy to say, not too confusing to hear, and suggests the ways in which we are about to try to use that triple.

The answer by Matt Samuel has already explained how to express the ordered triple in pure set notation. So I hope you can see now how words such as "equipped with two binary functions" can be parsed into more formal language; but aren't you glad we use "less formal" language to describe a ring instead of always describing it directly in the language of basic set theory?

• I agree with your last sentence. Actually, I feel the same when I am writing proofs, you really don't need to worry about the underlying formal logical cosntruct all the times.
– Kun
Mar 17, 2016 at 4:39

You can think of rings as "sets with extra structure." This extra structure on a ring $R$ is given by specifying binary addition and multiplication maps $R \times R \overset{+}{\to} R$ and $R \times R \overset{\times}{\to} R$ satisfying certain axioms (which also require the existence of distinguished additive and multiplicative identity elements $0$ and $1$.)

However, rings aren't sets in the same way that pairs of cats and dogs belonging to the set $\mathcal{Cats} \times \mathcal{Dogs}$ aren't cats. To any pair of a cat and dog, you can canonically associate a cat (by taking the cat in the pair.) In the same way, you can canonically associate to mathematical objects that are given as "sets with extra structure" their underlying sets.

A set is not ordered. $$\{a,b,c\}$$ and $$\{b,c,a\}$$ denote the same thing. The elements are "of the same type"*.

A tuple is ordered. $$(a,b,c)$$ differs from $$(b,c,a)$$ and the components can be of "different types"*. Their position distinguishes them.

The elements of a set and the components of a tuple can very well be sets or tuples (mixing allowed).

A ring is a triple (3-tuple) consisting of a set and two operators, with a number of specific "algebraic" properties.

*Understand this in an informal sense, "type" is not really defined.

Generally, my opinion is that you should cure yourself of the compulsion to do this.

Here is something that is certainly the case: you can devise a method of constructing a particular set such that the structure you want is recoverable from it – i.e. for every ring you can define a particular set and some set-theoretic operations on it that recover from it the underlying set of elements and each operation as a function.

In fact, there's lots of ways to construct such a set, most no better or worse than any other – this is your first clue that none should be considered "canonical" or "fundamental" or "what is really going on". The other, more damning clue, is that which one you pick (or picking none at all) has precisely no impact on the ring theory you do afterwards.

You can, if you like, just think of ringhood as an arity three (or five if you include zero and one) logical relation, where a set and a function and a function are related if between them they satisfy the ring axioms. But I tend to think that even this is not enlightening or informative.

• Not sure that the OP can appreciate this answer. And not sure it does address the elementary questions.
– user65203
Jul 16, 2020 at 7:41